Difference between revisions of "2018 AMC 10A Problems/Problem 9"
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− | All of the triangles in the diagram below are similar to | + | ==Problem== |
+ | |||
+ | All of the triangles in the diagram below are similar to isosceles triangle <math>ABC</math>, in which <math>AB=AC</math>. Each of the 7 smallest triangles has area 1, and <math>\triangle ABC</math> has area 40. What is the area of trapezoid <math>DBCE</math>? | ||
<asy> | <asy> | ||
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<math>\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24 </math> | <math>\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24 </math> | ||
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− | |||
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− | ===Solution 2 | + | ==Solution 1== |
+ | Let <math>x</math> be the area of <math>ADE</math>. Note that <math>x</math> is comprised of the <math>7</math> small isosceles triangles and a triangle similar to <math>ADE</math> with side length ratio <math>3:4</math> (so an area ratio of <math>9:16</math>). Thus, we have <cmath>x=7+\dfrac{9}{16}x.</cmath> This gives <math>x=16</math>, so the area of <math>DBCE=40-x=\boxed{(E) 24}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
Let the base length of the small triangle be <math>x</math>. Then, there is a triangle <math>ADE</math> encompassing the 7 small triangles and sharing the top angle with a base length of <math>4x</math>. Because the area is proportional to the square of the side, let the base <math>BC</math> be <math>\sqrt{40}x</math>. Then triangle <math>ADE</math> has an area of 16. So the area is <math>40 - 16 = \boxed{24}</math>. | Let the base length of the small triangle be <math>x</math>. Then, there is a triangle <math>ADE</math> encompassing the 7 small triangles and sharing the top angle with a base length of <math>4x</math>. Because the area is proportional to the square of the side, let the base <math>BC</math> be <math>\sqrt{40}x</math>. Then triangle <math>ADE</math> has an area of 16. So the area is <math>40 - 16 = \boxed{24}</math>. | ||
− | + | ==Solution 3== | |
Notice <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]</math>. | Notice <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]</math>. | ||
Let the base of the small triangles of area 1 be <math>x</math>, then the base length of <math>\Delta ADE=4x</math>. Notice, <math>\big(\frac{DE}{BC}\big)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}</math>, then <math>4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\big(\frac{4}{\sqrt{40}}\big)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]</math> | Let the base of the small triangles of area 1 be <math>x</math>, then the base length of <math>\Delta ADE=4x</math>. Notice, <math>\big(\frac{DE}{BC}\big)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}</math>, then <math>4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\big(\frac{4}{\sqrt{40}}\big)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]</math> | ||
− | Thus, <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\big(1-\frac{2}{5}\big)=\frac{3}{5}\cdot 40=\boxed{24}</math> | + | Thus, <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\big(1-\frac{2}{5}\big)=\frac{3}{5}\cdot 40=\boxed{24}.</math> |
− | Solution | + | <asy> |
+ | unitsize(5); | ||
+ | dot((0,0)); | ||
+ | dot((60,0)); | ||
+ | dot((50,10)); | ||
+ | dot((10,10)); | ||
+ | dot((30,30)); | ||
+ | draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); | ||
+ | draw((10,10)--(50,10)); | ||
+ | label("$B$",(0,0),SW); | ||
+ | label("$C$",(60,0),SE); | ||
+ | label("$E$",(50,10),E); | ||
+ | label("$D$",(10,10),W); | ||
+ | label("$A$",(30,30),N); | ||
+ | draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); | ||
+ | draw((15,15)--(45,15)); | ||
+ | </asy> | ||
+ | |||
+ | |||
+ | |||
+ | ==Solution 4== | ||
+ | The area of <math>ADE</math> is 16 times the area of the small triangle, as they are similar and their side ratio is <math>4:1</math>. Therefore the area of the trapezoid is <math>40-16=\boxed{24}</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be <math>7+5+3+1=16</math>, so to find the area of such trapezoid <math>BCED</math>, we just take <math>40-16=\boxed{24}</math>, like so. | ||
+ | |||
+ | ==Solution 6== | ||
+ | The combined area of the small triangles is <math>7</math>, and from the fact that each small triangle has an area of <math>1</math>, we can deduce that the larger triangle above has an area of <math>9</math> (as the sides of the triangles are in a proportion of <math>\frac{1}{3}</math>, so will their areas have a proportion that is the square of the proportion of their sides, or <math>\frac {1}{9}</math>). Thus, the combined area of the top triangle and the trapezoid immediately below is <math>7 + 9 = 16</math>. The area of trapezoid <math>BCED</math> is thus the area of triangle <math>ABC-16 =\boxed{24}</math>. | ||
+ | |||
+ | ==Solution 7== | ||
+ | You can assume for the base of one of the smaller triangles to be <math>\frac{1}{a}</math> and the height to be <math>2a</math>, giving an area of 1. The larger triangle above the 7 smaller ones then has base <math>\frac{3}{a}</math> and height <math>6a</math>, giving it an area of <math>9</math>. Then the area of triangle <math>ADE</math> is <math>16</math> and <math>40-16=\boxed{24}</math>. | ||
− | == See Also == | + | ==Video Solution== |
+ | https://youtu.be/ZiZVIMmo260 | ||
+ | |||
+ | https://youtu.be/tq4rWskQDh8 | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
{{AMC10 box|year=2018|ab=A|num-b=8|num-a=10}} | {{AMC10 box|year=2018|ab=A|num-b=8|num-a=10}} | ||
+ | {{AMC12 box|year=2018|ab=A|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 15:41, 22 July 2020
Contents
Problem
All of the triangles in the diagram below are similar to isosceles triangle , in which . Each of the 7 smallest triangles has area 1, and has area 40. What is the area of trapezoid ?
Solution 1
Let be the area of . Note that is comprised of the small isosceles triangles and a triangle similar to with side length ratio (so an area ratio of ). Thus, we have This gives , so the area of .
Solution 2
Let the base length of the small triangle be . Then, there is a triangle encompassing the 7 small triangles and sharing the top angle with a base length of . Because the area is proportional to the square of the side, let the base be . Then triangle has an area of 16. So the area is .
Solution 3
Notice . Let the base of the small triangles of area 1 be , then the base length of . Notice, , then Thus,
Solution 4
The area of is 16 times the area of the small triangle, as they are similar and their side ratio is . Therefore the area of the trapezoid is .
Solution 5
You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be , so to find the area of such trapezoid , we just take , like so.
Solution 6
The combined area of the small triangles is , and from the fact that each small triangle has an area of , we can deduce that the larger triangle above has an area of (as the sides of the triangles are in a proportion of , so will their areas have a proportion that is the square of the proportion of their sides, or ). Thus, the combined area of the top triangle and the trapezoid immediately below is . The area of trapezoid is thus the area of triangle .
Solution 7
You can assume for the base of one of the smaller triangles to be and the height to be , giving an area of 1. The larger triangle above the 7 smaller ones then has base and height , giving it an area of . Then the area of triangle is and .
Video Solution
~savannahsolver
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.